BZOJ 1406 密码箱(数论)

很简洁的题目。求出x^2%n=1的所有x<=n的值。 n<=2e9.

直接枚举x一定是超时的。 看看能不能化成有性质的式子。

有 (x+1)(x-1)%n==0，设n=a*b,那么一定有x+1=k1a,x-1=k2b. 不妨设a<=b.那么就能O(sqrt(n))枚举a。

然后再枚举x，验证x是否满足这两个式子。注意不能令x=k1a-1.由于a比较小，枚举x=k2b+1,k2b-1即可。

另外set很好用啊。

# include <cstdio>

# include <cstring>

# include <cstdlib>

# include <iostream>

# include <vector>

# include <queue>

# include <stack>

# include <map>

# include <set>

# include <cmath>

# include <algorithm>

using namespace std;

# define lowbit(x) ((x)&(-x))

# define pi 3.1415926535

# define eps 1e-

# define MOD

# define INF

# define mem(a,b) memset(a,b,sizeof(a))

# define FOR(i,a,n) for(int i=a; i<=n; ++i)

# define FO(i,a,n) for(int i=a; i<n; ++i)

# define bug puts("H");

# define lch p<<,l,mid

# define rch p<<|,mid+,r

# define mp make_pair

# define pb push_back

typedef pair<int,int> PII;

typedef vector<int> VI;

# pragma comment(linker, "/STACK:1024000000,1024000000")

typedef long long LL;

int Scan() {

    int res=, flag=;

    char ch;

    if((ch=getchar())=='-') flag=;

    else if(ch>=''&&ch<='') res=ch-'';

    while((ch=getchar())>=''&&ch<='')  res=res*+(ch-'');

    return flag?-res:res;

}

void Out(int a) {

    if(a<) {putchar('-'); a=-a;}

    if(a>=) Out(a/);

    putchar(a%+'');

}

const int N=;

//Code begin...

set<LL>::iterator it;

set<LL>S;

int main ()

{

    LL a, b, x, n;

    scanf("%lld",&n);

    for (int i=; i*i<=n; ++i) {

        if (n%i) continue;

        a=i; b=n/i;

        for (int k=; (x=b*k+)<n; ++k) if ((x+)%a==) S.insert(x);

        for (int k=; (x=b*k-)<n; ++k) if ((x-)%a==) S.insert(x);

    }

    for (it=S.begin(); it!=S.end(); ++it) printf("%lld\n",*it);

    return ;

}